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Subject: Electricity (multiple choice questions)
Date: Wed, 29 Dec 2004 09:51:39 -0500
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<HTML><HEAD><TITLE>Electricity - Electrostatics II</TITLE>
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<BODY text=3D#000000 vLink=3D#800080 aLink=3D#ff0000 link=3D#0000ff =
bgColor=3Dwhite>
<BLOCKQUOTE>
  <HR color=3Dblue SIZE=3D5>
  <TABLE bgColor=3D#ffff00>
    <TBODY>
    <TR>
      <TD><B>Multiple Choice Online Test</B><BR>Indicate the correct =
answer by=20
        clicking on the button next to the numbers 1, 2, 3 or 4. There =
is only=20
        one correct answer per question. After you have answered=20
        <B><U>all</U></B> multiple choice questions compare your choices =
with=20
        the correct answers. </TD></TR></TBODY></TABLE>
  <HR width=3D"100%" color=3Dred SIZE=3D3>

  <CENTER>
  <H2>Electrostatics II</H2></CENTER>
  <HR width=3D"100%" color=3Dred SIZE=3D3>
  <BR>
  <FORM name=3DF><!frage1><B><FONT color=3Dred>1.)</FONT> A small =
negatively charged=20
  oil drop moves slowly downwards between two oppositely charged plates =
in a=20
  Millikan type experiment. <BR>Taking which one of the following =
measures will=20
  not slow the drop down? </B>
  <P>
  <TABLE cellSpacing=3D0 cellPadding=3D0 width=3D"90%" border=3D0>
    <TBODY>
    <TR>
      <TD vAlign=3Dtop>
        <UL>
          <TABLE cellSpacing=3D0 border=3D0>
            <TBODY>
            <TR>
              <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ1></TD>
              <TD>Making the charge on the droplet more =
negative</TD></TR>
            <TR>
              <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ1></TD>
              <TD>Decreasing the distance between the plates</TD></TR>
            <TR>
              <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ1></TD>
              <TD>Increasing the potential difference across the =
plates</TD></TR>
            <TR>
              <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ1></TD>
              <TD>Changing the polarity of the =
plates</TD></TR></TBODY></TABLE><!-- <input type=3D"button" =
value=3D"Explanation" onclick=3D'gibfeedback("The electric current =
splits up according to the resistance (low resistance high current and =
high resistance low current). If 3 A flow into the parallel connection =
2/3 of I (2/3 of 3A =3D 2A) will pass through the 4&Omega; resistor and =
1/3 of I (1/3 of 3A =3D 1A) through the 8&Omega; resistor.")'>
 -->
          <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
        </UL></TD>
      <TD vAlign=3Dtop align=3Dmiddle>&nbsp;&nbsp; <IMG height=3D101=20
        alt=3D"electrostatics II, diagram 1"=20
        src=3D"http://prkw.com/pmcestatb01.gif" width=3D200 =
border=3D0>=20
    </TD></TR></TBODY></TABLE>
  <P><!frage2><B><FONT color=3Dred>2.)</FONT> A length of wire is placed =
at right=20
  angles to a magnetic field as shown in the diagram below. The wire =
carries a=20
  current in the direction as shown by the arrow on the wire. </B>
  <CENTER><IMG height=3D169 alt=3D"electrostatics II, diagram 2"=20
  src=3D"http://prkw.com/pmcestatb02.gif" width=3D400=20
  border=3D0></CENTER><B>In what direction will a force be exerted on =
point P=20
  situated on the conductor as shown in the sketch? </B>
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ2></TD>
        <TD>Upwards (out of the plane of the paper)</TD>
        <TD>&nbsp;</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ2></TD>
        <TD>Downwards (into the plane of the paper)</TD>
        <TD>&nbsp;</TD></TR>
      <TR>
        <TD vAlign=3Dcenter width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio =
value=3D1=20
          name=3DQ2></TD>
        <TD vAlign=3Dcenter>At 90<SUP>0</SUP> to the wire on a bearing =
of=20
          315<SUP>0</SUP> (as shown in the diagram)</TD>
        <TD><IMG height=3D58 alt=3D"electrostatics II, diagram 2a"=20
          src=3D"http://prkw.com/pmcestatb02a.gif" =
width=3D100=20
        border=3D0></TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ2></TD>
        <TD>At 90<SUP>0</SUP> to the wire on a bearing of =
135<SUP>0</SUP> (as=20
          shown in the diagram)</TD>
        <TD><IMG height=3D58 alt=3D"electrostatics II, diagram 2b"=20
          src=3D"http://prkw.com/pmcestatb02b.gif" =
width=3D100=20
        border=3D0></TD></TR></TBODY></TABLE><!-- <input type=3D"button" =
value=3D"Explanation" onclick=3D'gibfeedback("Since the total resistance =
is the same (2&Omega; + 8&Omega; =3D 10&Omega;) in both parts of the =
parallel connection all ammeters will show the same reading.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL><!&Omega;>
  <P><!frage3><B><FONT color=3Dred>3.)</FONT> When an inflated balloon =
is rubbed=20
  against a sweater the balloon acquires a negative charge. The balloon =
is then=20
  placed against a vertical wooden door and sticks to it. What is the =
reason for=20
  this? </B>
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ3></TD>
        <TD>wooden doors naturally have a positive charge</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ3></TD>
        <TD>the negative charge on the balloon induces a positive charge =
on=20
          the surface of the door</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ3></TD>
        <TD>the surface of the door becomes negative by transfer of =
charge=20
          from the balloon</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ3></TD>
        <TD>the surface of the door becomes positive by transfer of =
charge=20
          from the balloon</TD></TR></TBODY></TABLE><!-- <input =
type=3D"button" value=3D"Explanation" =
onclick=3D'gibfeedback("Ohm&acute;s Law: the current flowing through a =
conductor is directly proportional to the potential difference across it =
provided the temperature of the conductor remains constant. Or in =
formulae form: R =3D I/V, to keep R constant I and V have to =
increase/decrease at the same ratio.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <P><!frage4><B><FONT color=3Dred>4.)</FONT> Two identical metal =
spheres P and Q=20
  are mounted on insulated stands. Sphere P carries a charge of +4 nC =
and sphere=20
  B a charge of -4 nC. The spheres are separated by a distance d and =
exert a=20
  force of magnitude F on each other. P and Q touch and are returned to =
half=20
  their original distance apart (d/2). What is the magnitude of the =
force they=20
  exert on each other? </B>
  <CENTER><IMG height=3D142 alt=3D"electrostatics II, diagram 3"=20
  src=3D"http://prkw.com/pmcestatb03.gif" width=3D537 =
border=3D0></CENTER>
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ4></TD>
        <TD align=3Dmiddle>0</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ4></TD>
        <TD align=3Dmiddle>F</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ4></TD>
        <TD align=3Dmiddle>2 F</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ4></TD>
        <TD align=3Dmiddle>4 F</TD></TR></TBODY></TABLE><!-- <input =
type=3D"button" value=3D"Explanation" onclick=3D'gibfeedback("Unsrewing =
one of the parallel bulbs will change the total resistance of the =
circuit in a way that it goes up. Higher resistance will cause a lower =
current to flow through the circuit (option 2 and 4). How is the p.d. =
affected? Prior to unscrewing the voltmeter measured the p.d. across one =
of the bulbs connected in parallel, which had a lower resistance since =
they were in parallel (1/R<sub>total</sub> =3D 1/R<sub>1</sub> + =
1/R<sub>2</sub>). After unscrewing a higher voltage is required to drive =
the current through the bulb &raquo; voltmeter reading goes up.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <P><!frage5><B><FONT color=3Dred>5.)</FONT> An electron is between two =

  oppositely charged parallel plates in a vacuum as shown in the sketch. =
Which=20
  one of the following statements about the acceleration is true? </B>
  <TABLE width=3D"90%" border=3D0>
    <TBODY>
    <TR>
      <TD vAlign=3Dtop><BR>
        <UL><B>The acceleration of the electron is =85</B>=20
          <P>
          <TABLE border=3D0>
            <TBODY>
            <TR>
              <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ5></TD>
              <TD>zero</TD></TR>
            <TR>
              <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ5></TD>
              <TD>greatest close to the positive plate</TD></TR>
            <TR>
              <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ5></TD>
              <TD>greatest close to the negative plate</TD></TR>
            <TR>
              <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ5></TD>
              <TD>not zero, and the same everywhere between the=20
            plates</TD></TR></TBODY></TABLE><!-- <input type=3D"button" =
value=3D"Explanation" onclick=3D'gibfeedback("No current will flow in =
the circuit due to the position of the voltmeter. Voltmeters have a very =
high resistance - ammeters on the other hand have a very low resistance =
- and hence we can say that the resistance of the voltmeter <i>blocks =
somehow the current</i>. Since the voltmeter is connected across the =
terminals of the source of electricity it will measure the emf of the =
battery.")'>
 -->
          <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
          </P></UL></TD>
      <TD vAlign=3Dtop>&nbsp;&nbsp; <IMG height=3D206=20
        alt=3D"e-statics II, diagram 4"=20
        src=3D"http://prkw.com/pmcestatb04.gif" width=3D250 =
border=3D0>=20
    </TD></TR></TBODY></TABLE><!&Omega;>
  <P><!frage6><B><FONT color=3Dred>6.)</FONT> Two identical sized =
conducting=20
  spheres A, with a charge of +3,2 x 10<SUP>-19</SUP>C, and B, with a =
charge of=20
  -6,4 x 10<SUP>-19</SUP>C, are allowed to touch. During touching, =
sphere A ...=20
  </B>
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ6></TD>
        <TD>gains 3 protons</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ6></TD>
        <TD>loses 3 protons</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ6></TD>
        <TD>gains 3 electrons</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ6></TD>
        <TD>gains 2 electrons</TD></TR></TBODY></TABLE><!-- <input =
type=3D"button" value=3D"Explanation" onclick=3D'gibfeedback("Only in =
option 2 the effective resistance is 6&Omega;: 4&Omega; and 4&Omega; in =
parallel - 1/R =3D 1/4 + 1/4 =3D 1/2 hence R =3D 2&Omega;. 4&Omega; is =
in series with 2&Omega; hence R combined =3D 2&Omega; + 4&Omega; =3D =
6&Omega;.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <P><!frage7><B><FONT color=3Dred>7.)</FONT> Three identical metal =
spheres L, M=20
  and N are suspended by non-conducting threads as shown. The charges on =
L and M=20
  are -1 nC and +5 nC respectively, while N is neutral. If M is first =
brought=20
  into contact with L and then with N, and separated again, the final =
charge on=20
  sphere N will be =85 </B>
  <TABLE cellSpacing=3D0 cellPadding=3D0 width=3D"80%" border=3D0>
    <TBODY>
    <TR>
      <TD>
        <UL>
          <TABLE cellSpacing=3D0 border=3D0>
            <TBODY>
            <TR>
              <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ7></TD>
              <TD align=3Dmiddle>+4 nC</TD></TR>
            <TR>
              <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ7></TD>
              <TD align=3Dmiddle>+2 nC</TD></TR>
            <TR>
              <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ7></TD>
              <TD align=3Dmiddle>+1,5 nC</TD></TR>
            <TR>
              <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ7></TD>
              <TD align=3Dmiddle>+1 nC</TD></TR></TBODY></TABLE><!-- =
<input type=3D"button" value=3D"Explanation" =
onclick=3D'gibfeedback("B<sub>2</sub> and B<sub>3</sub> are connected in =
parallel, hence only 50% of the current will flow through each of those =
bulbs and the p.d. across them will be less than the p.d. across =
B<sub>1</sub> (1/3 of total p.d. will be across B<sub>2</sub> and =
B<sub>3</sub>; 2/3 of total p.d. across B<sub>1</sub>). As a result =
B<sub>1</sub> will shine brightest (power =3D current <small>x</small> =
voltage) and B<sub>2</sub> and B<sub>3</sub> will have the same =
brightness. ")'>
 -->
          <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
        </UL></TD>
      <TD>&nbsp;&nbsp;&nbsp;&nbsp;<IMG height=3D175=20
        alt=3D"e-statics II, diagram 5"=20
        src=3D"http://prkw.com/pmcestatb05.gif" width=3D251=20
    border=3D0></TD></TR></TBODY></TABLE>
  <P><!frage8><B><FONT color=3Dred>8.)</FONT> Two point charges =
+Q<SUB>1</SUB> and=20
  +Q<SUB>2</SUB> are kept separated at a distance r. The force of =
repulsion=20
  between them is F. If each of the charges is doubled, the force of =
repulsion=20
  between the charges will be: </B>
  <UL>
    <TABLE cellSpacing=3D0 cols=3D2 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ8></TD>
        <TD align=3Dmiddle>16 F</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ8></TD>
        <TD align=3Dmiddle>8 F</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ8></TD>
        <TD align=3Dmiddle>4 F</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ8></TD>
        <TD align=3Dmiddle>2 F</TD></TR></TBODY></TABLE><!-- <input =
type=3D"button" value=3D"Explanation" onclick=3D'gibfeedback("Coulombs =
Law: F =3D k Q<sub>1</sub>&middot;Q<sub>2</sub> / r<sup>2</sup>. =
Doubling Q causes the electric force between the point charges to become =
4 times greater.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <P>
  <P><!frage9><B><FONT color=3Dred>9.)</FONT> The diagram below (a) =
shows two=20
  metal spheres, A and B, on insulating stands, touching each other.=20
  <CENTER><IMG height=3D174 alt=3D"e-statics II, diagram 6"=20
  src=3D"http://prkw.com/pmcestatb06.gif" width=3D600=20
  border=3D0></CENTER>A negatively charged rod is now brought close to A =
as in the=20
  diagram (b). While the negatively charged rod is held in the same =
position,=20
  sphere B is separated from sphere A, after which the negatively =
charged rod is=20
  removed (c). How will the charge on sphere B compare with the charge =
on sphere=20
  A?</B>=20
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ9></TD>
        <TD>sphere B will have a negative charge of equal =
magnitude</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ9></TD>
        <TD>sphere B will have a positive charge of equal =
magnitude</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ9></TD>
        <TD>sphere B will have less negative charge</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ9></TD>
        <TD>sphere B will have no charge</TD></TR></TBODY></TABLE><!-- =
<input type=3D"button" value=3D"Explanation" =
onclick=3D'gibfeedback("Correct answer is option 3 (see also question =
7): 1/3 of total p.d. will be across single resistor R and 2/3 of total =
p.d. will be measured across resistors R in parallel. <br>2/3 &middot; =
12V =3D 8V; <br>and <br>1/3 &middot; 12V =3D 4V across R in =
parallel.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <P><!frage10><B><FONT color=3Dred>10.)</FONT> Millikan's famous oil =
drop=20
  experiment showed that charge carried by oil drops was quantised. =
Apart from=20
  the mass of the oil drop, and knowledge of the acceleration due to =
gravity,=20
  which other quantities are required for this charge to be calculated? =
</B>
  <UL>
    <TABLE cellSpacing=3D0 border=3D0>
      <TBODY>
      <TR>
        <TD width=3D50><B>1</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ10></TD>
        <TD>the velocity of the oil drop</TD></TR>
      <TR>
        <TD width=3D50><B>2</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ10></TD>
        <TD>the voltage between the plates</TD></TR>
      <TR>
        <TD width=3D50><B>3</B>&nbsp;<INPUT type=3Dradio value=3D1 =
name=3DQ10></TD>
        <TD>the voltage between the plates and the distance between =
them</TD></TR>
      <TR>
        <TD width=3D50><B>4</B>&nbsp;<INPUT type=3Dradio value=3D0 =
name=3DQ10></TD>
        <TD>the voltage between the plates, the distance between them =
and the=20
          velocity of the oil drop</TD></TR></TBODY></TABLE><!-- <input =
type=3D"button" value=3D"Explanation" onclick=3D'gibfeedback("Total =
internal resistance =3D 6 &middot; 0,2&Omega; =3D 1,2&Omega;<br>Voltage =
=3D 9V<br> I =3D V/R =3D 9/1,2 =3D 7,5A.")'>
 -->
    <SCRIPT language=3DJavaScript>zaehle()</SCRIPT>
  </UL>
  <HR color=3Dblue SIZE=3D5>

  <CENTER>
  <TABLE cellSpacing=3D0 cols=3D1 width=3D"60%" border=3D1>
    <CAPTION><B>Results</B></CAPTION>
    <TBODY>
    <TR>
      <TD>
        <CENTER><INPUT type=3Dhidden value=3Dfalse =
name=3Dbearbeitet><INPUT onclick=3Dshow() type=3Dbutton value=3D"See =
your results" name=3DButton></CENTER></TD></TR>
    <TR>
      <TD>
        <CENTER><TEXTAREA name=3Dergebnis rows=3D3 readOnly =
cols=3D50></TEXTAREA></CENTER></TD></TR>
    <TR>
      <TD>
        <CENTER><INPUT onclick=3Dshowcorrect() type=3Dbutton =
value=3D"What are the correct answers?" =
name=3DButton></CENTER></TD></TR></TBODY></TABLE></CENTER></FORM>

  <HR color=3Dblue SIZE=3D5>
</CENTER></BLOCKQUOTE></BODY></HTML>

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